# Does a regular pair of elements in a noetherian domain remain regular if their order is switched?

Recall that in a commutative ring $A$ an ordered pair of elements (a,b) is said to form a regular sequence if the ideal $\langle a,b\rangle$ is strictly included in $A$ ,if $a$ is not a zero-divisor in $A$ and if the class of $b$ is not a zero-divisor of $A/\langle a\rangle$.
A friend of mine has asked me if in that case we can conclude that $\langle b,a\rangle$ is also a regular sequence under the assumption that $A$ is a noetherian domain.
The answer is known to be yes for a local noetherian ring $A$, even it is not a domain

[Since I couldn't answer his question, I suggested to my friend that he ask here but he prefers that I do that]

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The only part to be shown is that $a$ is not a zero-divisor on $A/(b)$. Consider some $s\in A$ such that $as\in(b)$, say $as=bt$. Since $b$ is not a zero-divisor on $A/(a)$, we conclude that $t$ maps to zero in $A/(a)$, i.e., $t=au$. Since $A$ is a domain, it follows that $s=bu$, so $s$ maps to zero in $A/(b)$.
Strange, my friend ( a great algebraist who definitely knows what a quotient ring is!) told me he knew a counter-example due to Dieudonné in the case $A$ was a non-noetherian domain. He must have misremembered. Anyway your answer is obviously correct and optimal: thanks a lot. – Georges Elencwajg Oct 27 '11 at 16:24
Why is this the only part to be shown? We also need to show that $b$ is not a zero-divisor on $A$. That, it seems to me, is the hard part. – Steven Landsburg Oct 27 '11 at 16:28